How do you solve Query Kth Smallest Trimmed Number on LeetCode?

Query Kth Smallest Trimmed Number (LeetCode #2343) can be solved using Brute Force. The optimal approach is Brute Force with O(n² log n) time complexity and O(n) space complexity. Key insight: The brute force approach involves trimming the numbers for each query, sorting them, and then finding the k-th smallest trimmed number. It's straightforward but inefficient for larger inputs.

What are the interview tips for Query Kth Smallest Trimmed Number?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as all numbers being the same or having leading zeros. Practice explaining your thought process as you code.

What are common mistakes when solving Query Kth Smallest Trimmed Number?

Not considering leading zeros in trimmed numbers. Overlooking the importance of original indices when numbers are equal.

Query Kth Smallest Trimmed Number — LeetCode #2343 (Medium)

Tags: Array, String, Divide and Conquer, Sorting, Heap (Priority Queue), Radix Sort, Quickselect

Related patterns: Sorting, Pairing

Brute Force approach

Time complexity: O(n² log n). Space complexity: O(n).

The brute force approach involves trimming the numbers for each query, sorting them, and then finding the k-th smallest trimmed number. It's straightforward but inefficient for larger inputs.

The complexity arises from sorting the trimmed numbers for each query, leading to O(n log n) for sorting and O(n) for creating the list, repeated for each query.

Step 1: For each query, trim each number in nums to the specified number of rightmost digits.
Step 2: Pair each trimmed number with its original index.
Step 3: Sort the pairs based on the trimmed numbers, and if they are equal, sort by the original index.
Step 4: Retrieve the index of the k-th smallest trimmed number from the sorted list.

For nums = ["102", "473", "251", "814"] and queries = [[1, 1], [2, 3], [4, 2], [1, 2]]: 1. Query [1, 1]: Trimmed = ["2", "3", "1", "4"] → Sorted = [("1", 2), ("2", 0), ("3", 1), ("4", 3)] → Result = 2 2. Query [2, 3]: Trimmed = ["102", "473", "251", "814"] → Sorted = [("102", 0), ("251", 2), ("473", 1), ("814", 3)] → Result = 2 3. Query [4, 2]: Trimmed = ["02", "73", "51", "14"] → Sorted = [("02", 0), ("14", 3), ("51", 2), ("73", 1)] → Result = 1 4. Query [1, 2]: Trimmed = ["02", "73", "51", "14"] → Sorted = [("02", 0), ("14", 3), ("51", 2), ("73", 1)] → Result = 0

Common Mistakes

Not considering leading zeros in trimmed numbers.
Overlooking the importance of original indices when numbers are equal.

Interview Tips

Always clarify the problem requirements before jumping into coding.
Think about edge cases, such as all numbers being the same or having leading zeros.
Practice explaining your thought process as you code.

#2343

Query Kth Smallest Trimmed Number

Medium

Array↗String↗Divide and Conquer↗Sorting↗Heap (Priority Queue)↗Radix Sort↗Quickselect↗Sorting↗Pairing↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n² log n)Space O(n)

The brute force approach involves trimming the numbers for each query, sorting them, and then finding the k-th smallest trimmed number. It's straightforward but inefficient for larger inputs.

⚙️

Algorithm

4 steps

1Step 1: For each query, trim each number in nums to the specified number of rightmost digits.
2Step 2: Pair each trimmed number with its original index.
3Step 3: Sort the pairs based on the trimmed numbers, and if they are equal, sort by the original index.
4Step 4: Retrieve the index of the k-th smallest trimmed number from the sorted list.

solution.py7 lines

1def smallestTrimmedNumbers(nums, queries):
2    result = []
3    for k, trim in queries:
4        trimmed = [(num[-trim:], i) for i, num in enumerate(nums)]
5        trimmed.sort()
6        result.append(trimmed[k-1][1])
7    return result

ℹ

Complexity note: The complexity arises from sorting the trimmed numbers for each query, leading to O(n log n) for sorting and O(n) for creating the list, repeated for each query.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.